The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R - Physics MCQ AIEEE 2009 2026 | ExamTest.live

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The height at which the acceleration due to gravity becomes g /9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is:

Solution & Step-by-step Explanation

The formula for g at height h is g^{'} = g (\frac{R}{R + h})^{2} .Given g^{'} = g /9 : \frac{g}{9} = g (\frac{R}{R + h})^{2} \frac{1}{3} = \frac{R}{R + h} ⟹ R + h = 3 R ⟹ h = 2 R

Try it yourself before checking the explanation above.

The height at which the acceleration due to gravity becomes g /9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is:

2 R

R

\frac{R}{2}

2 R

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